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Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: x) dx=\frac{\pi}{8}\:ln(2)}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: x) dx \\ \\ \implies \bold{let} \: \: \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: x) dx = I \\ \\ \implies \bold{x = ( \frac{\pi}{4} - y) = > dx = - dy }\\ \\ \implies - \int \limits_{ \frac{\pi}{4} }^{ 0 } ln(1 + tan \: ( \frac{ \pi }{4} - y)) dy \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: (\frac{ \pi }{4} - y)) dy \\ \\ \bold {tan (A - B) = \frac{tan \: A - tan \: B}{1 + tan \:A \times tan \: B} } \\ \\ \implies tan( \frac{\pi}{4} - y) = \frac{tan \frac{\pi}{4} - tan \: y }{1 + tan \frac{\pi}{4} \times tan \: y} \\ \\ \bold { \implies tan( \frac{\pi}{4} - y) = \frac{1 - tan \: y}{1 + tan \: y} } \\ \\ \bold{ putting \: value \:of \: tan( \frac{\pi}{4} - y) } \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: (\frac{ 1 - tan \: y }{1 + tan \: y} )) dy$

$\implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + \: \frac{ 1 - tan \: y }{1 + tan \: y} ) dy \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln( \frac{1 + \cancel{ tan \: y} + 1 \cancel{- \: tan \: y}}{1 + \: tan \: y} ) dy \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln( \frac{2 }{1 + tan \: y} ) dy \\ \\ \bold{ln (\frac{A}{B} ) = ln \: A - ln \: B} \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ( ln(2) - ln(1 + tan \: y))dy \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(2) dy- \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: y)dy \\ \\ \implies \int \limits_{ 0 }^{ \frac{\pi}{4} } ( ln(2) dy- I$

$\implies ln(2) [y]_{0}^{ \frac{\pi}{4} } - I\\ \\ \implies ln(2)( \frac{\pi}{4} - 0) - I\\ \\ \implies I = \frac{ \pi }{4} \: ln(2) - I\\ \\ \implies 2 I = \frac{\pi}{4} \: ln(2) \\ \\ \implies I = \frac{\pi}{8} \: ln(2) \\ \\ \bold{\implies \frac{\pi}{8} \: ln(2)}$

${\bold{\therefore \int \limits_{ 0 }^{ \frac{\pi}{4} } ln(1 + tan \: x) dx=\frac{\pi}{8}\:ln(2)}}$

Answer 2

Step-by-step explanation:

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